Binomial Theorem Test

Result

Binomial Theorem Test - 41

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $$C_0, C_1, C_2, C_3, .... $$ are binomial coefficients in the expansion of $$ ( 1+x)^n $$ , then $$ \dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac{}{} - \dfrac{}{} + ... $$ is equal to :

A
$$ \dfrac{1}{n+1} - \dfrac{2}{n+2} + \dfrac{1}{n+3} $$

B
$$ \dfrac{1}{n+1} + \dfrac{2}{n+2} - \dfrac{3}{n+3} $$

C
$$ \dfrac{1}{n+2} - \dfrac{1}{n+1} + \dfrac{1}{n+3} $$

D
$$ \dfrac{2}{n+1} - \dfrac{1}{n+2} + \dfrac{2}{n+3} $$

E
$$ \dfrac{1}{n+2} - \dfrac{2}{n+1} + \dfrac{3}{n+3} $$

Solution

We know $$ (1-x)^n = C_0 - C_1 x + C_2x^2 - ... + (-1)^n C_n \cdot x^n $$

On multiplying both sides bby $$x^2$$ , we get

$$ (1-x)^n x^2 = C_0x^2 - C_1 x^3 + C_2x^4 - ... + $$

On integrating both sides by taking limit $$0$$ to $$1$$.

Therefore, $$\displaystyle \int_0^1 ( 1-x)^n x^2 dx =\displaystyle \int_0^1 ( C_0x^2 - C_1x^3 + C_2x^4 + ....) dx$$

$$\displaystyle \int_0^1 x^n (1-x)^2 dx = \left[ C_0 \dfrac{x^3}{3} - C_1 \dfrac {x^4}{4} + C_2 \dfrac{x^5}{5} - ... \right]_0^1 $$

$$ \Rightarrow\displaystyle \int_0^1 x^n ( 1 + x^2 - 2x) dx = \dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac {C_2}{5} - ... $$

Here $$\dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac {C_2}{5} - ... $$

$$ = \left[ \dfrac{x^{n+1}}{n+1} + \dfrac{x^{n+3}}{n+3} - \dfrac{2x^{n+2}}{n+2} \right]_0^1 $$

$$ = \left[ \dfrac {1}{n+1} + \dfrac{1}{n+3} - \dfrac {2}{n+2} \right] $$
Question 2
1 / -0

The value of $$r$$ for which the coefficients of $$(r-5)$$th and $$(3r+1)$$th terms in the expansion of $${(1+x)}^{1/2}$$ are equal, is

A
$$4$$

B
$$9$$

C
$$12$$

D
None of these

Solution

Since, coefficient of $$(r-5$$th term$$=$$ coefficient of $$=(3r+1)$$th term
$$\quad { _{ }^{ 12 }{ C } }_{ r-6 }=\quad { _{ }^{ 12 }{ C } }_{ 3r }$$
$$\Rightarrow$$ $$r-6=3r$$
or $$12-r+6=3r$$
$$\rightarrow$$ $$2r=-6$$ or $$4r=18$$
$$\Rightarrow$$ $$r=-3$$ or $$r=\cfrac { 18 }{ 4 } $$
Hence no value of $$r$$ exist, because $$r$$ neither be negative nor in fraction
Question 3
1 / -0

If $$(1 + x + x^{2})^{n} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$$, then $$2a_{1} - 3a_{2} + ... -(2n + 1)a_{2n}$$ is equal to

A
$$n$$

B
$$-n$$

C
$$n + 1$$

D
$$-n - 1$$

E
$$-n + 1$$

Solution

Given,
$$(1 + x + x^{2})^{n} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$$

$$\Rightarrow x(1 + x + x^{2})^{n} = x + a_{1}x^{2} + a_{2}x^{3} + ... + a_{2n}x^{2n + 1}$$

On diffferentiating w.r.t. $$x$$, we get

$$(1 + x +x^{2})^{n} + x\cdot n(1 + x + x^{2})^{n - 1}(1 + 2x)$$$$= 1 + 2a_{1}x + 3a_{2}x^{2} + ... + a_{2n} \cdot (2n + 1)x^{2n}$$

On putting $$x = -1$$, we get

$$(1 - 1 + 1)^{n} - n(1 - 1 + 1)^{n - 1}(1 - 2)$$$$= 1 - 2a_{1} + 3a_{2} + ... + a_{2n} (2n + 1)$$

$$\Rightarrow 1 - n(-1) = 1 - 2a_{1} + 3a_{2} + ... + a_{2n}(2n + 1)$$

$$\Rightarrow 2a_{1} - 3a_{2} .... =(2n + 1)a_{2n} = -n$$
Question 4
1 / -0

The middle term in the expansion of $$\left( \dfrac{10}{x} + \dfrac{x}{10} \right )^{10}$$ is

A
$$^{10}C_5$$

B
$$^{10}C_6$$

C
$$^{10}C_5 \dfrac{1}{x^{10}}$$

D
$$^{10}C_5 x^{10}$$

E
$$^{10}C_5 10^{10}$$

Solution

MIddle term of the expansion of $$\left(\cfrac{10}x+\cfrac x{10}\right)^{10}$$ is $$T_6.$$
$$T_6={}^{10}C_{5}\left(\cfrac{10}x\right)^5\left(\cfrac x{10}\right)^5$$
$$T_6={}^{10}C_{5}$$
Hence, A is the correct option.
Question 5
1 / -0

If the term free from $$x$$ in the expansion of $$\left (\sqrt {x} - \dfrac {k}{x^{2}}\right )^{10}$$ is $$405$$, then the value of $$k$$ is

A
$$\pm 1$$

B
$$\pm 3$$

C
$$\pm 4$$

D
$$\pm 2$$

Solution

General term in the expansion of $$\left (\sqrt {x} - \dfrac {k}{x^{2}}\right )$$

$$T_{r + 1} = ^{10}C_{r} (\sqrt {x})^{10 - r} \left (\tfrac {-k}{x^{2}}\right )^{r}$$

$$= ^{10}C_{r}x^{\dfrac {10 - r}{2}} \cdot (-k)^{r} \cdot x^{-2r}$$

$$= ^{10}C_{r} (-k)^{r} x^{\left (\dfrac {10 - 5r}{2}\right )}$$

The term is free from $$x$$.

Put $$\dfrac {10 - 5r}{2} = 0$$

$$\Rightarrow r = 2$$

Now, $$^{10}C_{2} (-k)^{2} = 405$$

$$\Rightarrow \dfrac {10\times 9}{1\times 2}\cdot k^{2} = 405$$

$$\Rightarrow k^{2} = \dfrac {405}{45} = 9$$

$$\Rightarrow k = \pm 3$$.
Question 6
1 / -0

If $${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },\dots ,{ C }_{ 15 }$$ are binomial coefficients in $${ \left( 1+x \right) }^{ 15 }$$, then $$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\dfrac { { C }_{ 3 } }{ { C }_{ 2 } } +\cdots +15\dfrac { { C }_{ 15 } }{ { C }_{ 14 } } $$ is equal to

A
$$60$$

B
$$120$$

C
$$64$$

D
$$124$$

E
$$144$$

Solution

We know that,
$$\dfrac { ^{ n }{ { C }_{ r } } }{ ^{ n }{ { C }_{ r-1 } } } =\dfrac { n-\left( r-1 \right) }{ r } $$

$$\Rightarrow r\cdot \dfrac { ^{ n }{ { C }_{ r } } }{ ^{ n }{ { C }_{ r-1 } } } =n+1-r$$

$$\Rightarrow \displaystyle\sum _{ r=1 }^{ n }{ r\cdot \dfrac { ^{ 16 }{ { C }_{ r } } }{ ^{ 16 }{ { C }_{ r-1 } } } } =\displaystyle\sum _{ r=1 }^{ 16 }{ \left( 16-r \right) } $$

$$\displaystyle 16\times 15-\sum_{r=1}^{n}r$$

$$=16\times 15-\dfrac { 15\times 16 }{ 2 } $$

$$=240-120$$

$$=120$$
Question 7
1 / -0

Sum of coefficients of the last $$6$$ terms in the expansion of $${ \left( 1+x \right) }^{ 11 }$$ when the expansion is in ascending powers of $$x$$, is

A
$$2048$$

B
$$32$$

C
$$512$$

D
$$64$$

E
$$1024$$

Solution

$$(1+x)^{\prime \prime}=\sum_{r=0}^{11} n_{C_{r}}(1)^{n-r}(x)^{r} \\$$

$$\text { last } 6 \text { terms of this expansion are } \\$$

$$\qquad{ }^{\prime \prime} C_{11}+{ }^{\prime \prime} C_{10}+{ }^{\prime \prime} C_{q}+{ }^{\prime \prime} C_{8}+{ }^{\prime \prime} C_{7}+{ }^{\prime \prime} C_{6} \\$$

$$\text { using }{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}$$

$$=12 C_{11}+{ }^{12} C_{9}+{ }^{12} C_{7} \\$$

$$=12+220+792 \\$$

$$=1024$$
Question 8
1 / -0

The middle term in the expansion of $${ \left( 1+x \right) }^{ 2n }$$ is

A
$$\cfrac { 1.3.5....(2n-1){ 2 }^{ n } }{ n! } $$

B
$$\cfrac { 1.2.3....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! } $$

C
$$\cfrac { 1.3.5....(2n-1){ x }^{ n } }{ n! } $$

D
$$\cfrac { 1.3.5....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! } $$

Solution

Given expansion is $${ \left( 1+x \right) }^{ 2n }$$,
The total number of terms are $$2n+1$$ which is an odd number,
$$\therefore$$ there is only one middle term which is $$\left( \begin{matrix} 2n \\ n \end{matrix} \right) $$,
$$\left( \begin{matrix} 2n \\ n \end{matrix} \right) =\dfrac { \left( 2n \right) ! }{ n!n! } \\ $$
$$\Longrightarrow \dfrac { \left[ 1.3.5...\left( 2n-1 \right) \right] \left[ { 2 }^{ n }n! \right] }{ n!n! } { x }^{ n }=\dfrac { \left[ 1.3.5...\left( 2n-1 \right) \right] \left[ { 2 }^{ n } \right] }{ n! } { x }^{ n }$$
Question 9
1 / -0

The value of $$\begin{pmatrix} 50 \\ 0 \end{pmatrix}\, \begin{pmatrix} 50 \\ 1 \end{pmatrix} + \begin{pmatrix} 50 \\ 1 \end{pmatrix}\, \begin{pmatrix} 50 \\ 2 \end{pmatrix}+........+\begin{pmatrix} 50 \\ 49 \end{pmatrix}\, \begin{pmatrix} 50 \\ 50 \end{pmatrix}$$ is , where $$^nC_r=\begin{pmatrix} n \\ r \end{pmatrix}$$

A
$$\begin{pmatrix} 100 \\ 50 \end{pmatrix}$$

B
$$\begin{pmatrix} 100 \\ 51 \end{pmatrix}$$

C
$$\begin{pmatrix} 50 \\ 25 \end{pmatrix}$$

D
$$\begin{pmatrix} 50 \\ 25 \end{pmatrix}^2$$

Solution

$$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+......+^nC_nx^n$$.
Now, $$(1+x)^{2n}=(1+x)^n.(1+x)^n$$
$$=\left(^nC_0+^nC_1x+^nC_2x^2+......^nC_{n-1}x^{n-1}+^nC_nx^n\right).\left(^nC_nx^n+^nC_{n-1}x^{n-1}+......+^nC_2x^2+^nC_1x+^nC_0\right)$$.
Now, equating the co-efficient of $$x^{n-1}$$ both sides we get,
$$^{2n}C_{n-1}=(^nC_0)(^nC_{n-1})+(^nC_1)(^nC_{n-2})+......(^nC_{n-1})(^nC_0)$$
or, $$^{2n}C_{n+1}=(^nC_0)(^nC_{1})+(^nC_1)(^nC_{2})+......(^nC_{n-1})(^nC_n)$$. {Since $$^nC_r=^nC_{n-r}$$]
Now, putting $$n=50$$ we get ,
$$^{100}C_{51}=$$$$(^{50}C_0)(^{50}C_{1})+(^{50}C_1)(^{50}C_{2})+......(^{50}C_{49})(^{50}C_{50})$$.
Question 10
1 / -0

If $$n\epsilon N$$ and $$(1 + 4x + 4x^{2})^{n} = \displaystyle \sum_{r = 0}^{r = 2n} a_{r}x^{r}$$ then value of $$2\displaystyle \sum_{r = 0}^{n}a_{2r}$$ equals

A
$$9^{n} - 1$$

B
$$9^{n} + 1$$

C
$$3^{n }+ 1$$

D
$$3^{n} - 1$$

Solution

$$\because 1 + 4x + 4x^{2} = (1 + 2x)^{2}$$
$$\therefore (1 + 2x)^{2n} = \displaystyle \sum_{r = 0}^{2n}a_{r}x^{r}$$
putting $$x = 1$$
$$3^{2n} = \displaystyle \sum_{r = 0}^{2n} a_{r} = a_{0} + a_{1} + a_{2} + a_{3} + ..... + a_{2n} .... (i)$$
and putting $$x = -1$$
$$1 = \displaystyle \sum_{r = 0}^{r = 2n}a_{r} (-1)^{r} = a_{0} - a_{1} + a_{2} - a_{3} + .... + a_{2n} ... (ii)$$
Now adding (i) and (ii) we get
$$2(a_{0} + a_{2} + a_{4} + .... + a_{2n}) = 3^{2n} + 1$$
$$\Rightarrow 2\displaystyle \sum_{r = 0}^{r =n} a_{2r} = 9^{n} + 1$$
Hence choice (b) is correct answer.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Binomial Theorem Test - 41

Result

Binomial Theorem Test - 41

Score

-

Rank

-

SHARING IS CARING

Question 1

$$ \dfrac{1}{n+1} - \dfrac{2}{n+2} + \dfrac{1}{n+3} $$

$$ \dfrac{1}{n+1} + \dfrac{2}{n+2} - \dfrac{3}{n+3} $$

$$ \dfrac{1}{n+2} - \dfrac{1}{n+1} + \dfrac{1}{n+3} $$

$$ \dfrac{2}{n+1} - \dfrac{1}{n+2} + \dfrac{2}{n+3} $$

$$ \dfrac{1}{n+2} - \dfrac{2}{n+1} + \dfrac{3}{n+3} $$

Solution

Question 2

$$4$$

$$9$$

$$12$$

None of these

Solution

Question 3

$$n$$

$$-n$$

$$n + 1$$

$$-n - 1$$

$$-n + 1$$

Solution

Question 4

$$^{10}C_5$$

$$^{10}C_6$$

$$^{10}C_5 \dfrac{1}{x^{10}}$$

$$^{10}C_5 x^{10}$$

$$^{10}C_5 10^{10}$$

Solution

Question 5

$$\pm 1$$

$$\pm 3$$

$$\pm 4$$

$$\pm 2$$

Solution

Question 6

$$60$$

$$120$$

$$64$$

$$124$$

$$144$$

Solution

Question 7

$$2048$$

$$32$$

$$512$$

$$64$$

$$1024$$

Solution

Question 8

$$\cfrac { 1.3.5....(2n-1){ 2 }^{ n } }{ n! } $$

$$\cfrac { 1.2.3....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! } $$

$$\cfrac { 1.3.5....(2n-1){ x }^{ n } }{ n! } $$

$$\cfrac { 1.3.5....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! } $$

Solution

Question 9

$$\begin{pmatrix} 100 \\ 50 \end{pmatrix}$$

$$\begin{pmatrix} 100 \\ 51 \end{pmatrix}$$

$$\begin{pmatrix} 50 \\ 25 \end{pmatrix}$$

$$\begin{pmatrix} 50 \\ 25 \end{pmatrix}^2$$

Solution

Question 10

$$9^{n} - 1$$

$$9^{n} + 1$$

$$3^{n }+ 1$$

$$3^{n} - 1$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-